Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MAXN = 1e5 + 10;
int n, q;
vector<int> G[MAXN];
int heavy[MAXN], sz[MAXN], p[MAXN], depth[MAXN];
vector<int> t[4 * MAXN];
vector<int> s[MAXN];
vector<int> arr[MAXN];
void build(int v, int b, int e) {
  if (b == e) {
    for (auto x : arr[b]) {
      t[v].push_back(x);
    }
    return;
  }
  int mid = (b + e) / 2;
  build((v << 1), b, mid);
  build((v << 1 | 1), mid + 1, e);
  t[v].resize(t[2 * v].size() + t[2 * v + 1].size());
  merge((t[2 * v]).begin(), (t[2 * v]).end(), (t[2 * v + 1]).begin(),
        (t[2 * v + 1]).end(), t[v].begin());
}
vector<int> add(const vector<int> &x, const vector<int> &y, int a) {
  vector<int> res;
  res.resize(x.size() + y.size());
  merge((x).begin(), (x).end(), (y).begin(), (y).end(), res.begin());
  if (a < res.size()) {
    res.resize(a);
  }
  return res;
}
vector<int> query(int v, int b, int e, int l, int r, int a) {
  if (b >= l && e <= r) {
    vector<int> res;
    int sz = t[v].size();
    for (int i = 0; i < min(sz, a); i++) res.push_back(t[v][i]);
    return res;
  }
  int mid = (b + e) / 2;
  if (mid >= r) {
    return query((v << 1), b, mid, l, r, a);
  }
  if (mid < l) {
    return query((v << 1 | 1), mid + 1, e, l, r, a);
  }
  vector<int> q1 = query((v << 1), b, mid, l, r, a);
  vector<int> q2 = query((v << 1 | 1), mid + 1, e, l, r, a);
  vector<int> res = add(q1, q2, a);
  return res;
}
void dfs(int u, int par = -1) {
  p[u] = par;
  sz[u] = 1;
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v != par) {
      depth[v] = depth[u] + 1;
      dfs(v, u);
      sz[u] += sz[v];
      if (!heavy[u] || sz[v] > sz[heavy[u]]) heavy[u] = v;
    }
  }
}
int head[MAXN], chainpos[MAXN], chain[MAXN], ch;
void hld() {
  dfs(1);
  for (int i = 1, j = 0; i <= n; ++i)
    if (p[i] == -1 || heavy[p[i]] != i) {
      ++ch;
      for (int k = i; k; k = heavy[k]) {
        arr[j] = s[k];
        chainpos[k] = j++, head[k] = i, chain[k] = ch;
      }
    }
  build(1, 0, n - 1);
}
vector<int> solve(int a, int b, int maxi) {
  vector<int> res;
  while (chain[a] != chain[b]) {
    if (depth[head[a]] < depth[head[b]]) swap(a, b);
    int start = head[a];
    res =
        add(res, query(1, 0, n - 1, chainpos[start], chainpos[a], maxi), maxi);
    a = p[start];
  }
  if (depth[a] > depth[b]) swap(a, b);
  res = add(res, query(1, 0, n - 1, chainpos[a], chainpos[b], maxi), maxi);
  return res;
}
int m;
int main() {
  scanf("%d%d%d", &n, &m, &q);
  int a, b, w;
  for (int i = 0; i < n - 1; i++) {
    scanf("%d%d", &a, &b);
    G[a].push_back(b);
    G[b].push_back(a);
  }
  for (int i = 0; i < m; i++) {
    int x;
    scanf("%d", &x);
    s[x].push_back(i + 1);
  }
  hld();
  int maxi;
  for (; q; q--) {
    scanf("%d%d%d", &a, &b, &maxi);
    vector<int> res = solve(a, b, maxi);
    printf("%d ", res.size());
    for (auto v : res) printf("%d ", v);
    printf("\n");
  }
  return 0;
}